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(2/5)y=4
We move all terms to the left:
(2/5)y-(4)=0
Domain of the equation: 5)y!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
(+2/5)y-4=0
We multiply parentheses
2y^2-4=0
a = 2; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·2·(-4)
Δ = 32
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*2}=\frac{0-4\sqrt{2}}{4} =-\frac{4\sqrt{2}}{4} =-\sqrt{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*2}=\frac{0+4\sqrt{2}}{4} =\frac{4\sqrt{2}}{4} =\sqrt{2} $
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